ELECTROLYSIS OF NaCl

The figure below shows an idealized drawing of a cell in which an aqueous solution of sodium chloride is electrolyzed.

Once again, the Na+ ions migrate toward the negative electrode and the Cl- ions migrate toward the positive electrode. But, now there are two substances that can be reduced at the cathode: Na+ ions and water molecules.

Cathode (-):

Na+ + e- → Na

Eored = -2.71 V

2 H2O + 2 e- → H2 + 2 OH-

Eored = -0.83 V

Because it is much easier to reduce water than Na+ ions, the only product formed at the cathode is hydrogen gas.

FARADAY'S LAW

The amount of a substance consumed or produced at one of the electrodes in an electrolytic cell is directly proportional to the amount of electricity that passes through the cell.

In order to use Faraday's law we need to recognize the relationship between current, time, and the amount of electric charge that flows through a circuit.

1 F = The quantity of electricity that is capable of depositing or liberating 1 gram equivalent weight( supply or react with one mole of electrons in a redox reaction) of a substance in electrolysis, approximately 9.6494 × 104 coulombs.

By definition, one coulomb of charge is transferred when a 1-amp current flows for 1 second.

(Around 6.242 × 1018electrons passing a given point each second constitutes one amp)

in other words: when of 6.242 × 1018 negative electrons pass a particular point in an electrolytic cell, that represents one coulomb...remember I do not mean 6.242 × 1018 x -1 C/electron, I mean 6.242 × 1018 x 1.602 x 10-19 C/electron.)

these electrons will come from an external power source...a battary.

1 C = 1 amp-s = 6.242 × 1018 electrons/point

and

each electron = 1.602 x 10-19 C

Example: To illustrate how Faraday's law can be used, let's calculate the number of grams of sodium metal that will form at the cathode when a 10.0-amp current is passed through molten sodium chloride for a period of 4.00 hours.

1. We start by calculating the amount of electric charge, in C, that flows through the cell.

10 C/sec x 3600 x 4 = 144,000 C/sec

Before we can use this information,

2. we need a bridge between this macroscopic quantity and the phenomenon that occurs on the atomic scale.

This bridge is represented by Faraday's constant, which describes the number of coulombs of charge carried by a mole of electrons.

Effect of concentration on Cell EMF

The copper electrode becomes lighter as copper atoms are oxidized to Cu2+ ions, which go into solution.

The silver electrode becomes heavier as Ag+ ions in the solution are reduced to silver metal.

The concentration of Cu2+ ions at the anode increases and the concentration of the Ag+ ions at the cathode decreases.

Negative ions flow from the salt bridge toward the anode to balance the charge on the Cu2+ ions produced at this electrode.

Positive ions flow from the salt bridge toward the cathode to compensate for the Ag+ ions consumed in the reaction.

An important property of the cell is missing from this list.

Over a period of time, the cell runs down, and eventually has to be replaced.

Let's assume that our cell is initially a standard-state cell in which the concentrations of the Cu2+ and Ag+ ions are both 1 molar.

Cu

|

Cu2+(1.0 M)

||

Ag+(1.0 M)

|

Ag

As the reaction goes forward copper metal is consumed and silver metal is produced therefore, the driving force behind the reaction must become weaker.

Therefore, the cell potential must become smaller.

Ecell = Eocell - (0.0257/n) ln Q

or in terms of log10

Ecell = Eocell - (0.0592/n) log Q

In the Nernst equation,

E is the cell potential at some moment in time,

Eo is the cell potential when the reaction is at standard-state conditions,

R is the ideal gas constant in units of joules per mole,

T is the temperature in kelvin,

n is the number of moles of electrons transferred in the balanced equation for the reaction,

F is the charge on a mole of electrons,

Qc is the reaction quotient at that moment in time.

The symbol ln indicates a natural logarithm, which is the log to the base e, where e is an irrational number equal to 2.71828...

Three terms in the Nernst equation are constants: R, T, and F.

The ideal gas constant is 8.314 J/mol-K. The temperature is usually 25oC.

Substituting this information into the Nernst equation gives the following equation.

Example: The standard-state potential for the Daniell cell is 0.46V. Two moles of electrons are transferred from copper metal to Ag+ ions in the balanced equation for this reaction, so n is 2 for this cell.

Because we never include the concentrations of solids in either reaction quotient or equilibrium constant expressions, Qc for this reaction is equal to the concentration of the Cu2+ ion divided by the concentration of the Ag+ ion.

Substituting what we know about the Daniell cell into the Nernst equation gives the following result, which represents the cell potential for the Daniell cell at 25oC at any moment in time.

The concentration of aqueous solutions in the half cells directly impact the EMF of the cell.

E = Eo - (0.0592 V/n) logk

Suppose we have this reaction:

Fe(s) + Cd2+(aq) ------> Fe2+(aq) + Cd(s)

In this reaction iron (Fe) is being oxidized in a 1.0 M solution of to iron(II) ion, while the cadmium ion (Cd2+) in a 0.5 M aqueous solution is being reduced to cadmium solid.

a. The first thing to answer is how does it behave in standard conditions?

We need to look at the standard potential for each half-reaction, then combine them to get a net potential for the reaction.

Notice that both half-reactions are shown as reductions -- the species gains electrons, and is changed to a new form.

To get the potential for the entire reaction, we add up the two (2) half-reactions to get 0.04 V for the standard potential.

E = Eo - (0.0592 V/n) logQ

E = 0.04 V - (0.0592 V/2) log(1.0/0.5)

E = 0.031 V

B. Suppose we now have a concentration of Cd2+ of 0.005 M, what is its potential?

E = 0.04 V - (0.0592 V/2) log(1.0/0.005)

E = 0.028 V

YOU TRY THIS ONE:

2 Ag+(aq) (0.80 M) + Hg(l) ---> 2 Ag(s) + Hg2+(aq) (0.0010M)

Answer: 0.025 V.

Since the value is positive, the reaction will work to form the products indicated.

Negative values of the potential indicate that the reaction tends to stay as reactants and not form the products.

Sample Calculation

Calculate the cell potential for the following system:

Write the half-reactions with the half-cell potentials:

Multiply the reactions to get the lowest common multiple of electrons:

When the reaction quotient is very small, the cell potential is positive and relatively large. This isn't surprising, because the reaction is far from equilibrium and the driving force behind the reaction should be relatively large.

When the reaction quotient is very large, the cell potential is negative. This means that the reaction would have to shift back toward the reactants to reach equilibrium.

Practice Problem

What is the potential of a cell made up of zinc and copper half-cells at the concentrations below:

Zn

|

Zn2+(0.25 M)

||

Cu2+(0.15M)

|

Cu

The cell potential depends on the logarithm of the ratio of the concentrations of the products and the reactants.

As a result, the potential of a cell or battery is more or less constant until virtually all of the reactants have been converted into products.

The Nernst equation can be used to calculate the potential of a cell that operates at non-standard-state conditions.

The sixth step involves multiplying each half-reaction by the smallest whole number that is required to equalize the number of electrons gained by reduction with the number of electrons produced by oxidation.

Using this guideline, the oxidation half reaction must be multiplied by "3" to give the 6 electrons required by the reduction half-reaction.

(ox.) 3 C2H6O ---> 3 C2H4O + 6 H+ + 6e-

The seventh and last step involves adding the two half reactions and reducing to the smallest whole number by cancelling species which on both sides of the arrow.

6e- + 14 H+ + (Cr2O7)-2 -----> 2 Cr+3 + 7 H2O

3 C2H6O -----> 3 C2H4O + 6 H+ + 6e-

adding the two half-reactions above gives the following:

6e-+ 14H++ (Cr2O7)-2 + 3C2H6O --> 2Cr+3 + 7H2O + 3C2H4O + 6H+ + 6e-

Note that the above equation can be further simplified by subtracting out 6 e- and 6 H+ ions from both sides of the equation to give the final equation.

Note: the equation above is completely balanced in terms of having an equal number of atoms as well as charges.

Voltaic Cells

1. Balance the following redox reactions, calculate Eo and predict whether each of the reactions will occur as written. Label the anode and cathode half reactions. The reactions are in acidic solution at standard conditions.

2. Given the following:

a) Write the shorthand notation for the cell that uses these half reactions.

b) Write the net equation for the cell (Hint: cancel spectators)

c) Calculate Eo for the cell

d) Calculate for the cell

2b. Use the standard reduction potentials listed below to calculate the standard free energy change ΔGo, for the following reaction.

4Ag(s) + O2(g) + 4H+(aq)→ 4Ag+(aq) + 2H2O(l)

Ox: O2(g) + 4H+(aq)→ 2H2O(l)Eo = +1.23 V

Red: 4Ag(s) → 4Ag+(aq) Eo = +0.80 V

a) Calculate Eo for the cell

b) Calculate for the cell

3. a) Use the electrode potentials to calculate the emf of a standard cell that uses the reaction:

b) What is for the cell?

c) if for the reaction is -93.09 kJ what is for the reaction?

hint: ΔGo = ΔH -TΔS

T = 298 K (from STAP, standard ambient temperature and pressure)

4. Use the standard electrode potentials to calculate the equilibrium constant at 25°C for the reaction:

5.

a) Calculate the emf of a cell formed from a Zn2+(aq)/Zn(s) half reaction in

which [Zn2+] = 0.0500M and Cl2(g)/Cl-(aq) half reaction in which [Cl-] =

0.0500M and the pressure of Cl2(g) is 1.25 atm

b) Write the equation for the cell reaction.

c) Which electrode is negative?

6. What is the concentration of Cd2+(aq) in the cell:

Zn|Zn2+(aq, 0.0900M)||Cd2+(aq, ?M)|Cd(s)

if the emf of the cell is 0.400V?

7. An acidic solution containing Pb2+ ions is electrolyzed and PbO2(s) plated out on the anode.

(a) Write the chemical equation for the anode reaction.

(b) If a current of 0.750 A is used for 25.0 min., what mass of PbO2(s) plates out?

(c) If a solution contains 2.50g of Pb2+, how many minutes will it take to plate out all of the lead, as PbO2(s) using a current of 0.750 amps?

8. How many minutes will it take to plate out 6.00 g of Cd from a Cd2+ solution using a current of 6.00A?

9. For the cell:

U(s)|U3+(aq)||Ag+(aq)|Ag(s)

Eo is 2.588 V. Use the emf of the cell and Eo for the Ag+/Ag reaction to calculate Eo for the U3+/U half reaction.

Practice Problem 14:

Determine the oxidation number of the chromium in an unknown salt if electrolysis of a molten sample of this salt for 1.50 hours with a 10.0-amp current deposits 9.71 grams of chromium metal at the cathode.

Electrochemical Cell (Regents)

Calculate the standard cell potential produced by a voltaic cell consisting of a nickel electrode in contact with a solution of Ni2+ ions and a silver electrode in contact with a solution of Ag+ ions.

What is the voltage produced by a voltaic cell consisting of an aluminum electrode in contact with a solution of Al3+ ions and an iron electrode in contact with a solution of Fe2+ ion?

Calculate the standard cell potential produced by a voltaic cell consisting of a sodium electrode in contact with a solution of Na+ ions and a copper electrode in contact with a solution of Cu2+ ions.

What is the voltage produced by a voltaic cell consisting of a calcium electrode in contact with a solution of Cu2+ ions

A voltaic cell is constructed using electrodes based on the following half reactions:

Pb2+ (aq) + 2e- -> Pb(s)

Au3+(aq) +3e- -> Au(s0

a. Which is the anode and which is the cathode in this cell?

b. What is the standard cell potential?

Calculate the standard cell potential produced by a voltaic cell consisting of a nickel electrode in contact with a solution of Ni2+ ions and a copper electrode in contact with a solution of Cu2+ ions

A voltaic cell is constructed using electrodes based on the following half reactions:

Mn2+ (aq) + 2e- -> Mn(s)

Cu2+ (aq) +2e- -> Cu(s)

Which is the anode and which is the cathode in this cell?

What is the standard cell potential?

What is the voltage produced by a voltaic cell consisting of a lead electrode in contact with a solution of Pb2+ ions and an iron electrode in contact with a solution of Fe2+? Which is anode and which is the cathode?

What is the voltage produced by a voltaic cell consisting of a zinc electrode in contact with a solution of Zn2+ ions and a silver electrode in contact with a solution of Ag+ ions?

Calculate the standard cell potential produced by a voltaic cell consisting of a gold electrode in contact with a solution of Au3+ ions and a silver electrode in contact with a solution of Ag+ ions. Which is the anode and which is the cathode?

Simple Voltaic Cell Practice

Analysis

1. Label the diagrams below to show the three identified voltaic cells. Identify electrodes, electrolytes, electron flow and direction of ion movement. Label the cathode, anode, and write the net cell reaction equations for each. (note: you decide which metals make up the electrodes)

2. Indicate whether the following processes occur at the cathode or at the anode of a voltaic cell.

a. reduction half-reaction _____________________________________

b. oxidation half-reaction _____________________________________

c. reaction of the strongest reducing agent ______________________

d. reaction of the strongest oxidizing agent ______________________

3. Use the tool to design the voltaic cell with the greatest cell potential. Explain why you chose this combination.